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Drupady [299]
2 years ago
15

Precal unit 4 sample work-

Mathematics
1 answer:
miv72 [106K]2 years ago
4 0

Answer:

  (3√26, 0)

Step-by-step explanation:

Finish your work by evaluating the sine and cosine. The imaginary part corresponds to the y-coordinate in (x, y) form.

  3√26(cos(360°) +i·sin(360°)) = 3√26(1 +0i)

In rectangular form, that is ...

  (3√26, 0)

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A number that will exceed the unknown number<br> by <br> 5
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Step-by-step explanation:

click on the image for the full answer

6 0
3 years ago
Jason rowed 1/3 of a mile down the river.Jane rowed 2/6 of a mile down the river. Jude rowed 2/9 of a mile down the river.What i
seraphim [82]

Answer:

8 / 9 miles

Step-by-step explanation:

Given that:

Jason's distance = 1/3 mile

Jane's distance = 2/6 miles

Jude's distance = 2/9 miles

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3 years ago
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ahrayia [7]
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5 0
3 years ago
(a) Show that a differentiable function f decreases most rapidly at x in the direction opposite the gradient vector, that is, in
Sophie [7]

Answer:

Step-by-step explanation:

\text{Show that a differentiable function f decreases most rapidly at x in the }

\text{direction opposite the gradient vector, that is, in the direction of} -\bigtriangledown f(x)\text{. Let}\  \theta \ \text{be the angle between} \bigtriangledown f(x) \  \text{and unit vector u. Then } D_u f = \mathbf{|\bigtriangledown f| \  cos  \ \theta }}

\text{Since the minimum value of} \ \  \mathbf{cos   \ \theta} \  \ is \mathbf{-1} \  \text{occuring \ for \ 0} \le \ \theta \ < 2x,  \\ \\ when  \ \theta = \mathbf{\pi} , \text{the mnimum value of} \  D_uf  \ is} -|\bigtriangledown f|,  \text{occuring when the direction of u is } \\ \\  \ \mathbf{the \ opposite \  of} \  \text{the direction of }  \ \bigtriangledown f (assuming \ \bigtriangledown f\ is \  not \ zero)

b) \text{From part A:}

If \ f(x,y) = x^4y -x^2y^2 \ \  decreases \ fastest \ at \ the \point \ (2,-5)\\ \\ F(x,y) = x^4y -x^2y^3 \\ \\ f_x = \dfrac{df}{dx}= \dfrac{d}{dx}(x^4y-x^2y^3)  \\ \\ f_x = \dfrac{df}{dx}= y4x^3 -2y^3x  \\ \\ For(2,-5) \\ \\ f_x = (-5)4(2)^3 -2(-5)^3(2) \\ \\ \mathbf{ f_x = 340}

However; f_y = \dfrac{df}{dy} = \dfrac{d}{dy}(x^4y - x^2y^3) \\ \\ f_y = x^4 -3x^2y^2 \\ \\  Now, for (2, -5)\\ \\f_y = (2)^4 -3(2)^2(-5)^2 \\ \\ f_y = -284

So; \bigtriangledown = < 340,-284> \text{this is the direction of fastest decrease}

6 0
3 years ago
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